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\title{\heiti\zihao{2} 习题7.1}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{已知下列定积分都存在,利用定积分的定义计算下列积分:}
\subsection{$\int_{0}^{1} c \mathrm{~d} x$}
\textbf{解}\quad
$$\int_{0}^{1} c \mathrm{~d} x=\lim _{\lambda(\Delta_{n}) \rightarrow 0} \sum_{i=1}^{n} c \Delta x{i}=c$$

\subsection{$\int_{0}^{1} x \mathrm{~d} x$}
\textbf{解}\quad
$$\int_{0}^{1} x \mathrm{~d} x=\lim _{\lambda\left(\Delta_{n}\right) \rightarrow 0} \sum_{i=1}^{n} \frac{i}{n} \cdot \Delta x{i}=\lim _{n \rightarrow \infty} \frac{1}{n^{2}} \cdot \frac{n(n+1)}{2}=\frac{1}{2}$$

\subsection{$\int_{a}^{b} \frac{1}{x^{2}} \mathrm{~d} x,(b>a>0)$}
\textbf{解}$1^{\circ}$\quad
令 $t=\frac{1}{x},$ 有 $x=\frac{1}{t}, \int_{a}^{b} \frac{1}{x^{2}} \mathrm{~d} x=\int_{\frac{1}{a}}^{\frac{1}{b}} t^{2}\left(-\frac{1}{t^{2}}\right) \mathrm{d} t=\int_{\frac{1}{b}}^{\frac{1}{a}} 1 \mathrm{~d} t .$ 
取划分
$\frac{1}{b}<\frac{1}{b}+\frac{1}{n}\left(\frac{1}{a}-\frac{1}{b}\right)<\frac{1}{b}+\frac{2}{n}\left(\frac{1}{a}-\frac{1}{b}\right)<\cdots<\frac{1}{b}+\frac{n-1}{n}\left(\frac{1}{a}-\frac{1}{b}\right)<\frac{1}{a}$
及
$$
\xi_{i}=\frac{1}{b}+\frac{i}{n}\left(\frac{1}{a}-\frac{1}{b}\right), \quad i=1,2, \cdots, n
$$

有 $\Delta x_{i}=\frac{1}{n}\left(\frac{1}{a}-\frac{1}{b}\right),$ 则 $\sum_{i=1}^{n} f\left(\xi_{i}\right) \Delta x_{i}=\sum_{i=1}^{n} 1 \cdot \frac{1}{n}\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{1}{a}-\frac{1}{b}(n \rightarrow \infty),$ 即
$$
\int_{a}^{b} \frac{1}{x^{2}} \mathrm{~d} x=\frac{1}{a}-\frac{1}{b}
$$

\textbf{解}$2^{\circ}$\quad
取划分$a<a+\frac{1}{n}(b-a)<a+\frac{2}{n}(b-a)<\cdots<a+\frac{n-1}{n}(b-a)<b$

及
$$
\xi_{i}=a+\frac{i}{n}(b-a), i=1,2, \cdots, n
$$

有$\Delta x_{i}=\frac{1}{n}(b-a)$,则$\sum_{i=1}^{n} f(\xi_{i})  \Delta x_{i}=\frac{a-b}{n} \sum_{i=1}^{n} \frac{1}{\left(a+\frac{i}{n}(b-a)\right)^{2}}=\frac{a-b}{n} \sum_{i=1}^{n} \frac{n}{(n a+i(b-a))^{2}}$

易知
$$
\begin{aligned}
    \frac{n}{n a+(b-a)}-\frac{n}{n b+(b-a)}&=\frac{a-b}{n} \sum_{i=1}^{n} \frac{n^2}{(n a+i(b-a))(n a+(i+1)(b-a))}\\&<\sum_{i=1}^{n} f\left(\xi_{i}\right) \Delta x_{i}\\&<\frac{a-b}{n} \sum_{i=1}^{n} \frac{n^2}{(n a+(i-1)(b-a))(n a+i(b-a))}\\&=\frac{1}{a}-\frac{1}{b}
\end{aligned}
$$

两侧同时对$n$取极限,由夹逼定理,其极限为$\frac{1}{a}-\frac{1}{b}$,即$\int_{a}^{b} \frac{1}{x^{2}} \mathrm{~d} x=\frac{1}{a}-\frac{1}{b}$

\section{利用积分的几何意义求下列积分}
\subsection{$\int_{-1}^{1} \sqrt{1-x^{2}} \mathrm{~d} x$}
\textbf{解}\quad
其表示圆心为$(0,0)$,半径为$1$的圆的上半部分的面积.即$\frac{\pi}{2}$

\subsection{$\int_{a}^{b}\left(x-\frac{a+b}{2}\right) \mathrm{d} x$}
\textbf{解}\quad
其表示过$(\frac{a+b}{2},0)$的斜率为$1$且左右端点到此点距离相同的线段的积分,显然左右两侧的三角形面积相同,所以积分为$0$







\end{document}
\subsection{}
\textbf{解}\quad

\subsection{}
\textbf{证}\quad

\textbf{\textcolor{red}{注}}\quad